Let $\boldsymbol{\xi} = (\xi_1,\xi_2,\xi_3)$ such that $\xi_i\geq 0$and $\xi_1+\xi_2+\xi_3 = 1$. Let $Y\sim N_3(\boldsymbol{\mu}(\boldsymbol{\xi}), \mathrm{\Sigma}(\boldsymbol{\xi}))$, where the components of the covariance matrix (which is singular in our case) are function of $\boldsymbol{\xi}$

\begin{equation} \boldsymbol{\mu}(\boldsymbol{\xi}) = \begin{bmatrix} (\mu_1-\mu_2)\sqrt{\frac{\xi_1\xi_2}{\xi_1+\xi_2}}\\ (\mu_1-\mu_3)\sqrt{\frac{\xi_1\xi_3}{\xi_1+\xi_3}}\\ (\mu_2-\mu_3)\sqrt{\frac{\xi_2\xi_3}{\xi_2+\xi_3}}\end{bmatrix}. \end{equation} and

\begin{equation} \mathrm{\Sigma}(\boldsymbol{\xi}) = \begin{bmatrix} 1 & \sqrt{\frac{\xi_2\xi_3}{(\xi_2+\xi_1)(\xi_3+\xi_1)}} & -\sqrt{\frac{\xi_1\xi_3}{(\xi_1+\xi_2)(\xi_3+\xi_2)}}\\ \sqrt{\frac{\xi_2\xi_3}{(\xi_2+\xi_1)(\xi_3+\xi_1)}} & 1 & \sqrt{\frac{\xi_1\xi_2}{(\xi_1+\xi_3)(\xi_2+\xi_3)}}\\ -\sqrt{\frac{\xi_1\xi_3}{(\xi_1+\xi_2)(\xi_3+\xi_2)}} & \sqrt{\frac{\xi_1\xi_2}{(\xi_1+\xi_3)(\xi_2+\xi_3)}} & 1 \end{bmatrix}. \end{equation}

Now suppose it is known that \begin{equation} \phi(\boldsymbol{\mu},\boldsymbol{\xi},a) = \int_{-a}^{a}\int_{-a}^{a}\int_{-a}^{a}f_{Y}(y)dy \end{equation}

where $f_{Y}(y)$ is the PDF of $Y$ and $a$ is some fixed constant. Let $\boldsymbol{\mu} = (\mu_1,\mu_2,\mu_3)$. It is numerically verified that $\phi(\boldsymbol{\mu},\boldsymbol{\xi},a)$ are equal for $\boldsymbol{\mu} = (-\delta,\delta,0)$, $\boldsymbol{\mu} = (-\delta,0,\delta)$, and $\boldsymbol{\mu} = (0,-\delta,\delta)$ when $\xi_1 = \xi_2 = \xi_3 = 1/3$. I need to prove that this is true only when $\xi_1 = \xi_2 = \xi_3 = 1/3$. (I have verified numerically that these CDFs are equal for $\xi_1 = \xi_2 = \xi_3 = 1/3$.)